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NCERT Exemplar Class 11 Maths Solutions Chapter 5 Complex Numbers and Quadratic Equations

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NCERT Exemplar Class 11 Maths Solutions Chapter 5 Complex Numbers and Quadratic Equations

Edited By Ravindra Pindel | Updated on Sep 12, 2022 05:49 PM IST

NCERT Exemplar Class 11 Maths solutions chapter 5 covers Complex numbers and its applications. Complex numbers are something which we have been studying over the years with any ‘real number’ being said to be a complex number, which can be denoted as a variable or alphabet in simpler terms. The other number, along with the real number, is a part of the complex number, is called the ‘imaginary number.’ Both of them appear together in a problem. Terms such as integer, conjugate, square root, polar, etc. will be studied in NCERT Exemplar Class 11 Maths solutions chapter 5, that will be utilised to find results for these complex number problems.

NCERT Exemplar Class 11 Maths Solutions Chapter 5: Exercise - 1.3

Question:1

For a positive integer n, find the value of (1 - i)^n (1-1/i)^n

Answer:

(1-i)^n (1-1/i)^n
=(1-i)^n (1+i )^n
=(1-i^2 )^n
=2^n

Question:2

Evaluate \sum_{i=1}^{13}(i^n+i^{n+1} ) where n \epsilonN.

Answer:

= \sum_{i=1}^{13}(i^n+i^{n+1} )
= \sum_{i=1}^{13}(1+i) i^n
=(1+i)(1+i^2+i^3+i^4....+i^{12}+i^{13} )
=(1+i) \frac{i(i^{13}-1)}{(i-1)}
=(1+i) i\frac{(i-1)}{(i-1)}
=(1+i)i
=i+i^2
=i-1

Question:3

If \frac{(1+i)}{(1-i)}^3-\frac{(1-i)}{(1+i)}^3= x+iythen find (x, y).

Answer:

Given \left(\frac{1+i}{1-i}\right)^3-\left(\frac{1-i}{1+i}\right)^3= x+iy
=\left(\frac{1+i}{1-i}\right)^3-\left(\frac{1-i}{1+i}\right)^3
=\left (\frac{1+2i+i^2}{1-i^2} \right )^{3}-\left (\frac{1-2i+i^2}{1-i^2} \right )^{3}
=(\frac{2i}{2})^3-(\frac{-2i}{2})^3
=i^3-(-i^3 )
=2i^3=0-2i
Thus,(x,y)=(0,-2)

Question:5

If \left(\frac{1-i}{1+i}\right)^{100}=a+ib then find (a, b).

Answer:

a+ib=\left(\frac{1-i}{1+i}\right)^{100}=\left [\frac{1-i}{1+i}*\frac{1-i}{1-i} \right ]^{100}
=\left [\frac{(1-i)^2}{1-i^2} \right ]^{100}
=\left (\frac{(1-2i+i^2)}{(1+1)} \right )^{100}
=\left (\frac {-2i}{2 }\right )^{100}
=(i^4 )^{25}=1
Hence,(a,b)=(1,0)

Question:6

If a=\cos \theta+i \sin\theta, find the value of\frac{(1+a)}{(1-a)} .

Answer:

a=\cos \theta+i \sin\theta
\frac{(1+a)}{(1-a)}=\frac{(1+cos\theta)+i \sin \theta}{(1-cos\theta)-isin\theta}

=\frac{\left ( 2\cos ^{2}\frac{\theta }{2} +i2\sin\frac{\theta }{2}\cos\frac{\theta }{2} \right )}{\left ( 2\sin ^{2}\frac{\theta }{2} -i2\sin\frac{\theta }{2}\cos\frac{\theta }{2} \right )}
=\frac{2\cos\frac{\theta }{2}\left ( \cos\frac{\theta }{2}+i\sin\frac{\theta }{2} \right )}{2\sin\frac{\theta }{2}\left ( \sin\frac{\theta }{2}-i\cos\frac{\theta }{2} \right )}
=\frac{i\cos\frac{\theta }{2}\left ( \cos\frac{\theta }{2}+i\sin\frac{\theta }{2} \right )}{\sin\frac{\theta }{2}\left ( i\sin\frac{\theta }{2}-i^2\cos\frac{\theta }{2} \right )}
=\frac{i\cos\frac{\theta }{2}\left ( \cos\frac{\theta }{2}+i\sin\frac{\theta }{2} \right )}{\sin\frac{\theta }{2}\left ( i\sin\frac{\theta }{2}+\cos\frac{\theta }{2} \right )}
=i\cot \frac{\theta }{2}

Question:7

If (1 + i)z = (1 - i) \bar{z}, then show that z= -i\bar{z}

Answer:

(1 + i)z = (1 - i) \bar{z},
z=\frac{1-i}{1+i}\bar{z}
Rationalising the denominator,
\frac{\left (1-i \right )\left (1-i \right )}{\left (1+i \right )\left (1-i \right )}\bar{z}
=\frac{\left ( 1-i \right )^{2}}{1-i^2}\bar{z}
=\frac{\left ( 1-2i+i^2 \right )}{1+1}\bar{z}
=\frac{\left ( 1-2i-1 \right )}{2}\bar{z}
=-i\bar{z}
Hence,Proved.

Question:8

Ifz = x + iy, then show that z\bar{z}+2(z+\bar{z})+b=0 where b?R, representing z in the complex plane is a circle.

Answer:

z=x+iy
\bar{z}=x-iy
Now,we\: \: also\: \: have\: \: ,z\bar{z }+2(z+\bar{z})+b=0
(x+iy)(x-iy)+2(x+iy+x-iy)+b=0
This equation is a equation of circle

Question:9

If the real part of \frac{(\bar{z}+2)}{(\bar{z}-1)} is 4, then show that the locus of the point representing z in the complex plane is a circle.

Answer:

Let \: \: z=x+iy \: \:
\frac{\bar{z}+2}{\bar{z}-1}= \frac{x-iy+2}{x-iy-1}
=\frac{\left [ \left ( x+2 \right )-iy \right ]\left [ \left ( x-1 \right )+iy \right ]}{\left [ \left ( x-1 \right )-iy \right ]\left [ \left ( x-1 \right )+iy \right ]}
=\frac{(x-1)(x+2)+y^2+i[(x+2)y-(x-1)y]}{(x-1)^2+y^2}
real \: \: part=4 \Rightarrow \frac{(x-1)(x+2)+y^2}{(x-1)^2+y^2}=4
x^2+x-2+y^2=4(x^2-2x+1+y^2 )
3x^2+3y^2-9x+6=0

The equation obtained is that of a circle. Hence, locus of z is a circle.

Question:10

Show that the complex number z, satisfying the condition arg\left ( \frac{z-1}{z+1} \right )=\frac{\pi}{4} lies on a circle.

Answer:

Let z=x+iy
arg\left ( \frac{z-1}{z+1} \right )=\frac{\pi}{4}
arg\left ( z-1 \right )-arg\left ( z+1 \right )=\frac{\pi}{4}
arg\left ( x+iy-1 \right )-arg\left ( x+iy+1 \right )=\frac{\pi}{4}
arg\left ( x-1+iy \right )-arg\left ( x+1+iy \right )=\frac{\pi}{4}
\tan^{-1}\frac{y}{x-1}-\tan^{-1}\frac{y}{x+1}=\frac{\pi}{4}
\tan^{-1}\frac{\frac{y}{x-1}-\frac{y}{x+1}}{1+\left (\frac{y}{x-1} \right )\left (\frac{y}{x+1} \right )}=\frac{\pi}{4}
\tan^{-1}\left(\frac{y\left ( x+1+-x+1 \right )}{x^2-1+y^2}\right) = \frac{\pi}{4}
\tan\frac{\pi}{4} = \frac{2y}{x^2+y^2-1}
x^2+y^2-1=2y
x^2+y^2-1-2y=0
The equation obtained represents the equation of a circle

Question:11

Solve that equation |z| = z + 1 + 2i.

Answer:

|z| = z + 1 + 2i
Substituting \: \: z=x+iy \: \: we \: \: get \: \: |x+iy|=x+iy+1+2i
|z|= \sqrt{(x^2+y^2 )} =(x+1)+i(y+2)
Comparing real and imaginary parts \sqrt{(x^2+y^2 )} =(x+1)
And0=y+2 , y=-2
Substituting the value of y in\sqrt{(x^2+y^2 )} =(x+1)
x^2+(-2)^2=(x+1)^2
x^2+4=x^2+2x+1

Hence, x=\frac{3}{2}
Hence,z=x+iy
=\frac{3}{2} -2i

Question:12

If |z + 1| = z + 2 (1 + i), then find z

Answer:

We have |z + 1| = z + 2 (1 + i)
Substituting z=x+iy we get x+iy+1=x+iy+2i+1
|z|= \sqrt{(x^2+y^2 ) }=\sqrt{(x+1)^2+y^2 } = (x+2)+i(y+2)
Comparing real and imaginary parts, \sqrt{(x+1)^2+y^2 } = (x+2)
and 0=y+2 ;y=-2
Substituting the value of y in \sqrt{(x+1)^2+y^2 } = (x+2)
(x+1)^2+(-2)^2=(x+2)^2
x^2+2x+1+4=x^2+4x+4
2x=1
Hence, x=\frac{1}{2}
Thus, z=x+iy=\frac{1}{2}-2i

Question:13

If arg (z - 1) = arg (z + 3i), then find x – 1 : y. where z = x + iy

Answer:

Given that arg (z - 1) = arg (z + 3i)
arg (x+iy-1)=arg(x+iy+3i)
arg(x-1+iy)=arg(x+i(y+3))
\tan^{-1}\frac{y}{x-1}=\tan^{-1}\frac{y+3}{x}
\frac{y}{x-1}=\frac{y+3}{x}
xy=xy-y+3x-3
3x-3=y
\frac{x-1}{y}=\frac{1}{3}

Question:14

Show that \left | \frac{z-2}{z-3} \right |=2 represents a circle. Find its centre and radius.

Answer:

\left | \frac{z-2}{z-3} \right |=2 Substituting z=x+iy, we get \left | \frac{x+iy-2}{x+iy-3} \right |=2
|x-2+iy|=2|x-3+iy|
\sqrt{(x-2)^2+y^2 }=2\sqrt{((x-3)^2+y^2}
x^2-4x+4+y^2=4(x^2-6x+9+y^2 )
3x^2+3y^2-20x+32=0
x^2+y^2-\frac{20}{3} x+\frac{32}{}3=0
(x-\frac{10}{3})^2+y^2+\frac{32}{3}-\frac{100}{9}=0
Thus, the centre of circle is (\frac{10}{3},0) and radius is 2/3

Question:15

If \frac{z-1}{z+1} is a purely imaginary number(z \neq -1), then find the value of |z|.

Answer:

Let z=x+iy
\frac{z-1}{z+1}=\frac{x+iy-1}{x+iy+1}
\frac{\left [ \left ( x-1 \right ) +iy\right ]\left [ \left ( x+1 \right )-iy \right ]}{\left [ \left ( x+1 \right ) +iy\right ]\left [ \left ( x+1 \right )-iy \right ]}
=\frac{(x-1)(x+1)+y^2+i[(x+1)y-(x-1)y]}{(x+1)^2+y^2 }
\frac{z-1}{z-1} is purely imaginary
\frac{(x-1)(x+1)+y^2}{(x+1)^2+y^2 }=0
x^2-1+y^2=0
x^2+y^2=1 Hence, |z|=1

Question:17

If |z1| = 1 (z1 ≠ –1) and z_2= \frac{z_1-1}{z_1+1} then show that the real part of z2 is zero

Answer:

Let z_1=x+iy |z_1 |= \sqrt{x^2+y^2} =1
z_2= \frac{z_1-1}{z_1+1}=\frac{x+iy-1}{x+iy+1}
=\frac{[(x-1)+iy][(x+1)-iy]}{[(x+1)+iy][(x+1)-iy] }
= \frac{(x-1)(x+1)+y^2+i[(x+1)y-(x-1)y]}{(x+1)^2+y^2}
\frac{x^2+y^2-1+2iy}{(x+1)^2+y^2 }=0
Since, x^2+y^2=1
\frac{1-1+2iy}{(x+1)^2+y^2}=\frac{0+2iy}{(x+1)^2+y^2}
Therefore, the real part of z_2 is zero

Question:18

If z1, z2 and z3, z4 are two pairs of conjugate complex numbers, then find arg\left ( \frac{z_1}{z_4} \right )+arg\left ( \frac{z_2}{z_3} \right )

Answer:

z_1 and z_2 are conjugate complex numbers.
The negative side of the real axis = r_1 (cos\theta-isin \theta)
=r_1\left ( \cos\left ( -\theta_1 \right )+i\sin\left (- \theta_1 \right ) \right )
Similarly, z_3=r_2\left ( \cos\left ( \theta_2 \right )-i\sin\left ( \theta_2 \right ) \right )
z_4=r_2\left ( \cos\left ( -\theta_2 \right )+i\sin\left (- \theta_2 \right ) \right )
arg \left ( \frac{z_1}{z_4} \right )+arg \left ( \frac{z_2}{z_3} \right )=arg(z_1)-arg(z_4)+ arg(z_2)- arg(z_3)??
=\theta_1-(-\theta_2)+(-\theta_1)-\theta_2=0

Question:19

If |z_1| = |z_2| =... = |z_n| = 1, then
Show that |z_1 + z_2 + z_3 +... + z_n| =|1/z_1 +1/z_2 +...+1/z_n |

Answer:

|z_1| = |z_2| =... = |z_n| = 1
|z_1|^2 = |z_2|^2 =... = |z_n|^2 = 1
z_1 \bar{z_1 } =z_2 \bar{z_2 }=...=z_n \bar{z_n }
z_1=\frac{1}{\bar{z_1}},z_2=\frac{1}{\bar{z_2}},...z_n=\frac{1}{\bar{z_n}}
Now, |z_1+z_2+z_3+...z_n | =\left | \frac{z_1\bar{z_1}}{\bar{z_1}}+\frac{z_2\bar{z_2}}{\bar{z_2}}+...+\frac{z_n\bar{z_n}}{\bar{z_n}} \right |
=\left | \frac{1}{\bar{z_1}}+\frac{1}{\bar{z_2}}+...+\frac{1}{\bar{z_n}} \right |
=\left | {\frac{1}{z_1}+\frac{1}{z_2}+...+\frac{1}{z_n}} \right |

Question:21

Solve the system of equations Re (z^2) = 0, |z| = 2.

Answer:

Re (z^2) = 0, |z| = 2.
Let\: \: z=x+iy
|z|= \sqrt{x^2+y^2 } =2
{x^2+y^2 } =4
z^2=x^2+2ixy-y^2
=(x^2-y^2 )+2ixy
Now,Re (z^2 )=0
x^2-y^2=0
x^2=y^2=2
x=y=\pm \sqrt{2}
Hence, z=x+iy=\pm\sqrt{2}\pm i\sqrt{2}
=\sqrt{2}+i\sqrt{2},\sqrt{2}-i\sqrt{2},-\sqrt{2}+i\sqrt{2} \: \: and\: \: -\sqrt{2}-i\sqrt{2}

Question:22

Find the complex number satisfying the equation z+\sqrt{2} |(z+1)|+i=0 .

Answer:

z+\sqrt{2} |(z+1)|+i=0\: \: \: \: ...(i)
Substituting z=x+iy we get x+iy+\sqrt2 |x+iy+1|+i=0
x+i(1+y)+\sqrt2 \left [ \sqrt{(x+1)^2+y^2 } \right ]=0
x+i(1+y)+\sqrt2 \left [ \sqrt{(x^2+2x+1+y^2 ) } \right ]=0
1+y = 0
y = -1
x+\sqrt2 \sqrt{x^2+2x+2}=0
\sqrt2 \sqrt{x^2+2x+2}=-x
2x^2+4x+4=x^2
x^2+4x+4=0
(x+2)^2=0
x=-2
Hence,z=x+iy=-2-i

Question:25

i) For any two complex numbers z1, z2 and any real numbers a, b, |az_1 - bz_2|^2 + |bz_1 + az_2|^2 =....

ii) The value of \sqrt{-25} * \sqrt{-9} is

iii) The number \frac{(1-i)^3}{1-i^3} is equal to .......

iv) The sum of the series i+i^2+i^3+...+i^{1000} upto 1000 terms is ...

v) Multiplicative inverse of 1 + i is ................

vi) If z_1 \ and \ z_2 are complex numbers such that z_1 + z_2 is a real number, then z_2 =

vii) arg (z)+arg (\bar{z} ) (\bar{z}\neq0) is ....

viii) If |z+4|\leq 3then the greatest and least values of |z+1| are ..... and .....

ix) if \frac{z-2}{z+2}=\frac{\pi}{6}then the locus of z is .......

x) If |z|=4 and arg (z) = \frac{5\pi}{6} then z=

Answer:

i) |az_1 - bz_2|^2 + |bz_1 + az_2|^2
=|az_1 |^2+|bz_2 |^2-2Re(az_1.b(\bar{z_2} ) +|bz_1 |^2+|az_2 |^2+2Re(az_1.b\bar{z_2} ) ?)
=|az_1 |^2+|bz_2 |^2+|bz_1 |^2+|az_2 |^2=(a^2+b^2 )(|z_1 |^2+|z_2 |^2 )
ii)\sqrt{-25}*\sqrt{-9}=5i*3i=15i^2=-15
iii) \frac{(1-i)^3}{1-i^3}=\frac{(1-i)^3}{(1-i)(1+i+i^2 )}
=\frac{(1-i)^3}{(1+i-1) }= \frac{1+i^2-2i}{}i
=\frac{1-1-2i}{i}=-\frac{2i}{i}=-2
iv) i+i^2+i^3+...+i^{1000}=0 \left [ \sum_{n=1}^{1000}i^n=0 \right ]
v) \frac{1}{1+i}=1*\frac{1-i}{(1+i)(1-i)} =\frac{1}{2} (1-i)
vi) Let z_1=x_1+iy_1 \: \: and \: \: z_2=x_2+iy_2
z_1+z_2=(x_1+x_2 )+i(y_1+y_2 )
If z_1+z_2 is real then y_1+y_2=0
y_1=-y_2
z_2=x_2-iy_1
z_2=x_1-iy_1 (\: \: when\: \: x_1=x_2 )
So, z_2=\bar{z_1}
vii) arg (z)+arg (\bar{z} )
If\: \: arg (z)= \theta , then \arg (\bar{z})=-\theta
\theta+(-\theta)=0
viii) |z+4|\leq 3
=|z+4-3|\leq|z+4|+|-3|
=|z+4-3|\leq3+3
=|z+4-3|\leq6
The gratest value is 6 and the least value of |z+1| is 0
ix)\frac{z-2}{z+2}=\frac{\pi}{6}
Let z=x+iy
\left |\frac{x+iy-2}{x+iy+2} \right |=\left |\frac{(x-2)+iy}{(x+2)+iy} \right |= \frac{\pi}{6}
6|(x-2)+iy|=\pi|(x+2)+iy|
6\sqrt{(x-2)^2+y^2 }=\pi\sqrt{(x+2)^2+y^2 }
36[x^2+4-4x+y^2 ]=\pi ^2[x^2+4+4x+y^2 ]
36x^2+144-144x+36y^2=\pi^2 x^2+4\pi^2+4\pi^2 x+\pi^2 y^2
(36-\pi^2 ) x^2+(36-\pi^2 )-(144+4\pi^2 )x+144-4\pi^2=0
which represents an equation of a circle
x) |z|=4 and arg (z) = \frac{5\pi}{6}
Let z=x+iy
|z|= \sqrt{x^2+y^2} =4
x^2+y^2=16
arg(z)=\tan^{-1}\frac{y}{x}=\frac{5\pi}{6}
\frac{y}{x}=\tan\left ( \pi-\frac{\pi}{6} \right )
=-\tan \frac{\pi}{6}=-\frac{1}{\sqrt{3}}
x=-\sqrt3 y
(-\sqrt3 y)^2+y^2=16
3y^2+y^2=16
4y^2=16
y^2=4
y=\pm2
x=-2\sqrt3
z=-2\sqrt3+2i

Question:26

State True of False for the following:

i) The order relation is defined on the set of complex numbers.

ii) Multiplication of a non-zero complex number by -i rotates the point about origin through a right angle in the anti-clockwise direction
iii) For any complex number z, the minimum value of |z| + |z – 11 is 1
iv) The locus represented by |z — 11= |z — i| is a line perpendicular to the join of the points (1,0) and (0, 1)
v) If z is a complex number such that z ≠ 0 and Re(z) = 0, then Im (z2) = 0
vi) The inequality |z – 4| < |z – 2| represents the region given by x > 3.
(vii) Let Z1 and Z2 be two complex numbers such that |z, + z2| = |z1 j + |z2|, then arg (z1 – z2) = 0.
(viii) 2 is not a complex number.

Answer:

(i) Comparison of two purely imaginary complex numbers is not possible. However, the two purely real complex number can be compared. So, it is false.

(ii) Let z= x+iy

z.i = (x+iy)i =xi= -y which rotates at angle of 180. So, it is ‘false’.

(iii) Let z= x+iy

|z|+|z-1|= \sqrt{x^2+y^2} +\sqrt{(x-1)^2+y^2}

The value of |z|+|z-1| is minimum when x=0,y=0 \, \: \: i.e.,1

Hence, it is true.

iv) Let z= x+iy

Given that \left | z-1 \right |=\left |z-i \right |

|x+iy-1|=|x+iy-i|

|(x-1)+iy|=|x-(1-y)i|

\sqrt{(x-1)^2+y^2}=\sqrt{x^2+(1-y)}^2

(x-1)^2+y^2=x^2+(1-y)^2

x^2-2x+1+y^2=x^2+1+y^2-2y

-2x+2y=0 x-y=0 which is a straight line slope=1

Now, equation of line through the point 1,0and 0,1

y-0=\frac{1-0}{0-1} (x-1)

y=-x+1 whose slope=-1

Multiplication of the slopes of two lines =-1*1=-1

So, they are perpendicular. Hence, it is true.

v)Let z= x+iy z\neq0 \: \: and \: \: Re(z)=0

Since, real part is 0 x=0, z=0+iy=iy

1m(z^2 )=y^2 i^2=-y^2which is real Hence, it is False.

vi) (z-4)<|z-2|

Let z= x+iy

|x+iy-4|<|x+iy-2|

|(x-4)+iy|<|(x-2)+iy|

\sqrt{(x-4)^2+y^2 }<\sqrt{(x-2)^2+y^2}

(x-4)^2+y^2<(x-2)^2+y^2

(x-4)^2<(x-2)

x^2+16-8x<x^2+4-4x

8x+4x<-16+4

-4x<-12

x>3 Hence, it is true.

vii) Let z_1=x_1+iy_1 and z_2=x_2+iy_2

|z_1+z_2 |=|z_1 |+|z_2 |

|x_1+iy_1+x_2+iy_2 |=|(x_1+iy_1 )+(x_2+y_2 i)|

= \sqrt{(x_1+x_2 )^2+(y_1+y_2 )^2 }

=\sqrt{x_1^2+y_1^2} +\sqrt{x_2^2+y_2^2 }

Squaring both sides, we get (x_1^2+x_2^2+2x_1 x_2+y_1^2+y_2^2+2y_1 y_2)

=x_1^2+y_1^2+x_2^2+y_2^2+2\sqrt{x_1^2 x_2^2+x_1^2 y_2^2+x_2^2 y_1^2+y_1^2 y_2^2 }

=2x_1 x_2+2y_1 y_2=\sqrt{x_1^2 x_2^2+x_1^2 y_2^2+x_2^2 y_1^2+y_1^2 y_2^2 }

=x_1 x_2+y_1 y_2=\sqrt{x_1^2 x_2^2+x_1^2 y_2^2+x_2^2 y_1^2+y_1^2 y_2^2 }

Again squaring on both sides we get x_1^2 x_2^2+x_1^2 y_2^2+2x_1 y_1 x_2 y_2=x_1^2 x_2^2+x_1^2 y_2^2+x_2^2 y_1^2+y_1^2 y_2^2 2x_1 y_1 x_2 y_2=x_1^2 y_2^2+x_2^2 y_1^2

x_1^2 y_1^2+x_2^2 y_1^2-2x_1 y_1 x_2 y_2=0

(x_1 y_2-x_2 y_1 )^2=0

x_1 y_2-x_2 y_1=0

x_1 y_2=x_2 y_1

\frac{x_1}{y_1}=\frac{x_2}{y_2}

\frac{y_1}{x_1}=\frac{y_2}{x_2}

arg(z_1)=arg(z_2)=0 Hence, it is false.

(viii) Since, every real number is a complex number. So, 2 is a complex number. Hence, it is false.

Question:27

COLUMN A

COLUMN B

a) The polar form of i+\sqrt3 is

i) Perpendicular bisector of segment joining (-2,0) and (2,0)

b)The amplitude of -1+\sqrt{-3} is

ii) On or outside the circle having centre at (0,-4) and radius 3.

c) If \left | z+2 \right |=\left | z-2 \right | then locus of z is

iii) 2\pi/3

d)If \left | z+2i \right |=\left | z-2i \right | then locus of z is

iv)Perpendicular bisector of segment joining (0,-2) and (0,2)

e) Region represented by \left | z+4i \right |\geq 3 is

v)2\left ( \cos\frac{\pi}{6}+i\sin\frac{\pi}{6} \right )

f) Region represented by \left | z+4 \right |\geq 3 is

vi) On or outside the circle having centre at (-4,0) and radius 3units.

g)Conjugate of \frac{1+2i}{1-i} lies in

vii) First Quadrant

h) Reciprocal of 1-i lies in

viii) Third Quadrant

Answer:

a)Given z=i+\sqrt3

Polar form of z =r[cos\theta+isin\theta]=i+\sqrt3.

r=\sqrt{3+1} =2

And \tan \alpha = \frac{1}{\sqrt3}

\alpha = \frac{\pi}{6}

Since x>0,y>0

Polar form of z =2[cos\frac{\pi}{6}+isin\frac{\pi}{6}]

b) Given that z=-1+\sqrt3=-1+\sqrt3 i

Here argument z =\tan^{-1}\left | \frac{\sqrt3}{-1} \right |= \tan^{-1}\sqrt{3}= \frac{\pi}{3}

So, \alpha = \frac{\pi}{3}

Since, x<0 and y>0 \theta=\pi-\alpha=\pi-\frac{\pi}{3}=\frac{2\pi}{3}

c) |z+2|=|z-2|

|x+iy+2|=|x+iy-2|

|(x-2)+iy|=|(x-2)+iy|

\sqrt{(x+2)^2+y^2} =\sqrt{(x-2)^2+y^2}

(x+2)^2+y^2=(x-2)^2+y^2

(x+2)^2=(x-2)^2

x^2+4+4x=x^2+4-4x

8x=0 , x=0

which represents equation of-y axis and is perpendicular to the line joining the points (-2,0) and (2,0)

d) |z+2i|=|z-2i|

(x+iy+2i)=|x+iy-2i|

|x+(y+2)i|=|x+(y-2)i|

\sqrt{x^2+(y+2)^2} =\sqrt{x^2+(y-2)^2 }

x^2+(y+2)^2=x^2+(y-2)^2

(y+2)^2=(y-2)^2

y^2+4+4y=y^2+4-4y

8y=0 ,y=0

which is the equation of x-axis and is perpendicular to the line segment joining (0,-2)and (0,2)

e) |z+4i|\geq3

|x+iy+4|\geq3

|x+(y+4)i|\geq3

\sqrt{x^2+(y+4)^2} \geq3

x^2+(y+4)^2\geq9

x^2+y^2+8y+16 \geq9

x^2+y^2+8y+7 \geq0 r=\sqrt{(4)^2-7}=3

which represents a circle on or outside having centre (0,-4)

f) |z+4|\leq3

Let z=x+iy Then, |x+iy+4|\leq3

|(x+4)+iy|\leq3

\sqrt{(x+4)^2+y^2 } \leq3

x^2+8x+16+y^2\leq9

x^2+y^2+8x+7 \leq0 which is a circle having centre (-4,0) and r=\sqrt{(4)^2-7}=\sqrt9=3 and is on the circle.

g) Let z=\frac{1+2i}{1-i}

\frac{1+2i}{1-i}* \frac{1+i}{1+i}=\frac{1+i+2i+2i^2}{1-i^2}

=\frac{1+i+2i-2}{1+1}= \frac{-1+3i}{2}

=\frac{1+i+2i-2}{1+1}= \frac{-1+3i}{2}

=-\frac{1}{2}+\frac{3}{2}i which lies in third quadrant..

h) Given that z=1-i

Reciprocal of z=\frac{1}{z}=\frac{1}{1-i}*\frac{1+i}{1+i}

=\frac{1+i}{1+i^2} which lies in first quadrant.

So,(a)-(v),(b)-(iii),(c)-(i),(d)-(iv),(e)-(ii),(f)-(vi),(g)-(viii),(h)-(vii)

Question:28

What is the conjugate of \frac{2-i}{\left (1-2i \right )^2}?

Answer:

\frac{2-i}{\left (1-2i \right )^2}= \frac{2-i}{1+4i^2-4i}
= \frac{2-i}{-3-4i}= \frac{2-i}{-3-4i}*\frac{-3+4i}{-3+4i}
=\frac{-6+8i+3i-4i^2}{9-16i^2}
\frac{-6+11i+4}{9-16i^2}=\frac{-2+11i}{9+16}
\frac{-2+11i}{25}=\frac{-2}{25}+\frac{11}{25}i
Hence, \bar{z}=-\frac{2}{25}-\frac{11}{25}i

Question:29

If |z_1| = |z_2|, is it necessary that z_1 = z_2?

Answer:

Let z_1=x_1+iy_1 \: \: and\: \: z_2=x_2+iy_2
|x_1+iy_1 |=|x_2+iy_2 |
= \sqrt{x_1^2+y_1^2} = \sqrt{x_2^2+y_2^2}
x_1^2+y_1^2=x_2^2+y_2^2
\\$Consider $ z_1 = 3 + 4i \\ z_2 = 4 + 3i
z_1\neq z_2 Hence, it is not necessary that z_1= z_2

Question:31

Find z if |z| = 4 and arg (z) = 5\pi/6..

Answer:

Polar form of z=r[cos \theta+i\sin \theta]
=4\left [ \cos \frac{5\pi}{6}+i\sin \frac{5\pi}{6} \right ]
=4\left [ \cos\left ( \pi-\frac{\pi}{6} \right )+i\sin \left ( \pi-\frac{\pi}{6} \right ) \right ]
=4\left [ -\cos\left (\frac{\pi}{6} \right )+i\sin \left (\frac{\pi}{6} \right ) \right ]
=4\left [-\frac{\sqrt3}{2}+i\frac{1}{2} \right ]
=-2\sqrt3+2i

Question:32

Find (1+i)\frac{(2+i)}{(3+i)}.

Answer:

\left | (1+i)\frac{(2+i)}{(3+i)}*\frac{3-i}{3-i} \right |=\left | (1+i).\frac{6-2i+3i-i^2}{9-i^2} \right |
=\left | \frac{(1+i)(7+i)}{9+1} \right |
=\left | \frac{7+i+7i+i^2}{9+1} \right |
=\left | \frac{7+i+7i+i^2}{10} \right |
=\left | \frac{7+8i-1}{10} \right |
=\left | \frac{6+8i}{10} \right |
=\left |\frac{3}{5}+\frac{4}{5}i \right |
=\sqrt{\left (\frac{3}{5} \right )^2+\left (\frac{4}{5} \right )^2}=1

Question:33

Find principal argument of (1+i\sqrt3)^2.

Answer:

\begin{array}{c} (1+i \sqrt{3})^{2}=1+i^{2} \cdot 3+2 \sqrt{3} i \\ =1-3+2 \sqrt{3} i=-2+2 \sqrt{3} i \\ \tan \alpha=\left|\frac{2 \sqrt{3}}{-2}\right| \\ \tan \alpha=|-\sqrt{3}|=\sqrt{3} \\ \tan \alpha=\tan \frac{\pi}{3} \\ \alpha=\frac{\pi}{3} \end{array}

Now Re(z)<0 and image(z)>0 arg(z)=\pi-\alpha=\pi-\frac{\pi}{3}=\frac{2\pi}{3}
\text {Hence, the principal arg}=\frac{2 \pi}{3}

Question:34

Where does z lie, if |\frac{z-5i}{z+5i}|=1.

Answer:

\begin{array}{c} \text { Let } z=x+i y \quad\left|\frac{x+y i-5 i}{x+i y+5 i}\right|=1 \\ |x+(y-5) i|=|x+(y+5) i| \\ x^{2}+(y-5)^{2}=x^{2}+(y+5)^{2} \\ (y-5)^{2}=(y+5)^{2} \\ \qquad \begin{array}{r} y^{2}+25-10 y=y^{2}+25+10 y \\ 20 y=0 \quad y=0 \end{array} \end{array}

Hence, z lies on x-axis i.e., real axis.


Question:35

\sin x + i \cos 2x and \cos x - i \sin 2x are conjugate to each other for:
A. x = n\pi

B. x = \left ( n+\frac{1}{2} \right )\frac{\pi}{2}

C. x = 0
D. No value of x

Answer:

The answer is the option (c).
\text{let }z=\sin x+i \cos 2 x \quad \bar{z}=\sin x-i \cos 2 x

But \: \: we \: \: are\: \: given\: \: that\: \: \bar{z}=\cos x-i \sin 2 x$ $$

\sin x-\operatorname{icos} 2 x=\cos x-i \sin 2x Comparing the real and imaginary part, we get

\sin \mathrm{x}=\cos \mathrm{x}$ \: \: and\: \: $\cos 2 \mathrm{x}=\sin 2 \mathrm{x}

\tan \mathrm{x}=1 \text { and } \tan 2 \mathrm{x}=1
\tan 2 \mathrm{x}= \frac{2\tan x }{1-\tan^2 x}=1
The above value is not satisfied by tan x = 1. Hence no value of x is possible.

Question:36

The real value of α for which the expression \frac{\left ( 1-i\sin\alpha \right )}{1+2i\sin\alpha} is purely real is:
A.\left ( n+1 \right )\frac{\pi}{2}
B.\left ( 2n+1 \right )\frac{\pi}{2}
C.n\pi
D. None of these

Answer:

The answer is the option (c).
Let z=\frac{\left ( 1-i\sin\alpha \right )}{1+2i\sin\alpha}
\begin{aligned} &\begin{array}{c} =\frac{(1-i\sin \alpha)(1-2 i \sin \alpha)}{(1+2 i \sin \alpha)(1-2 i \sin \alpha)} \\\\ =\frac{1-2 i \sin \alpha-i \sin \alpha+2 i^{2} \sin ^{2} \alpha}{(1)^{2}-(2 i \sin \alpha)^{2}} \\\\ =\frac{1-3 i \sin \alpha-2 \sin ^{2} \alpha}{1-4 i^{2} \sin ^{2} \alpha}=\frac{\left(1-2 \sin ^{2} \alpha\right)-3 i \sin \alpha}{1+4 \sin ^{2} \alpha} \\\\ \frac{-3 i \sin \alpha}{1+4 \sin ^{2} \alpha}=0 \\\\ \sin \alpha=0 \quad \operatorname{} \alpha=n \pi \end{array}\\ &\text { Hence, c is correct. } \end{aligned}

Question:37

If z = x + iy lies in the third quadrant, the \frac{\bar{z}}{z}also lies in the third quadrant if
A. x > y > 0
B. x < y < 0
C. y < x < 0
D. y > x > 0

Answer:

The answer is the option (b).
If z lies in the third quadrant,So,x<0 \: \: and\: \: y<0\: \: , z=x+iy
\begin{aligned} \frac{\bar{z}}{z} &=\frac{x-i y}{x+i y}=\frac{x-i y}{x+i y} * \frac{x-i y}{x-i y} \\ &=\frac{x^{2}+i^{2} y^{2}-2 x y i}{x^{2}-i^{2} y^{2}} \\ &=\frac{x^{2}-y^{2}}{x^{2}+y^{2}}-\frac{2 x y}{x^{2}+y^{2}} i \end{aligned}
When z lies in third quadrant then \frac{\bar{z}}{z} will also be in third quadrant.
\begin{array}{c} \frac{x^{2}-y^{2}}{x^{2}+y^{2}}<0 \text { and } \frac{2 x y}{x^{2}+y^{2}}>0 \\\\ x^{2}-y^{2}<0 \text { and } 2 x y>0 \\\\ x^{2}<y^{2} \text { and } x y>0 \\\\ \text { So, } x<y<0 \end{array}
Hence, b is correct..

Question:38

The value of (z + 3) (\bar{z}+3) is equivalent to
A.|z + 3|^2
B. |z – 3|
C.z^2 + 3
D. None of these

Answer:

The answer is the option (a).
Let\: \: z=x+iy \: \: So,\: \: (z+3)(\bar{z} +3)=(x+iy+3)(x-iy+3)
=\left [ \left ( x+3 \right ) +iy\right ]\left [ \left ( x+3 \right ) -iy\right ]
=(x+3)^2-y^2 i^2=(x+3)^2+y^2
=|x+3+iy|^2=|z+3|^2

Question:39

If \left ( \frac{1+i}{1-i} \right )^{x}=1 , then
A. x = 2n + 1
B. x = 4n
C. x = 2n
D.x = 4n + 1

Answer:

The answer is the option(b).
\begin{array}{c} \left(\frac{1+i}{1-i}\right)^{x}=1 \\\\ \left(\frac{(1+i)(1+i)}{(1-i)(1+i)}\right)^{x}=1 \\\\ \left(\frac{1-1+2 i}{1+1}\right)^{x}=1 \\\\ \left(\frac{2 i}{2}\right)^{x}=1(i)^{x}=1 \end{array}
x = 4 n, n \epsilon N
Hence, the correct option is

A real value of x satisfies the equation \frac{3-4 i x}{3+4 i x}=\alpha-i \beta (\alpha, \beta \in R) , if \alpha ^2 +\beta ^2 =
A. 1
B. –1
C. 2
D. –2

Answer:

The answer is the option (a).

\begin{aligned} &\text {Given that} \frac{3-4 i x}{3+4 i x}=\alpha-i \beta\\ &=\frac{3-4 i x}{3+4 i x} * \frac{3-4 i x}{3-4 i x}=\alpha-i \beta\\ &\frac{9-12 i x-12 i x+16 i^{2} x^{2}}{9-16 i^{2} x^{2}}=\alpha-i \beta\\ &=\frac{9-16 x^{2}}{9+16 x^{2}}-\frac{24 x}{9+16 x^{2}}=\alpha-i \beta \ldots \ldots .(i)\\ &\frac{9-16 x^{2}}{9+16 x^{2}}+\frac{24 x}{9+16 x^{2}}=\alpha+i \beta \ldots \ldots \ldots(ii)\\\end{aligned}
Multiplying eqn (i )and (ii )we get
\left (\frac{9-16 x^{2}}{9+16 x^{2}} \right )^{2}+\left (\frac{24 x}{9+16 x^{2}} \right )^{2}=\alpha^{2}+ \beta^{2}
\frac{81+256 x^{4}-288 x^{2}+576x^2}{\left(9+16 x^{2}\right)^{2}}=\alpha^{2}+\beta^{2}
\begin{aligned}\frac{81+256 x^{4}+288 x^{2}}{\left(9+16 x^{2}\right)^{2}}=\alpha^{2}+\beta^{2}\\ \frac{\left(9+16 x^{2}\right)^{2}}{\left(9+16 x^{2}\right)^{2}}=\alpha^{2}+\beta^{2}=1 \end{aligned}

Question:41

Which of the following is correct for any two complex numbers z_1\: \: and\: \: \: \: z_2 ?
A. |z_1 z_2| = |z_1||z_2|
B. arg (z_1z_2) = arg (z_1). Arg (z_2)
C. |z_1 + z_2| = |z_1|+ |z_2|
D. |z_1 + z_2| \geq |z_1| - |z_2|

Answer:

The answer is the option (a).
\begin{array}{c} z_{1}=r_{1}\left(\cos \theta_{1}+i \sin \theta_{2}\right) \\ \left|z_{1}\right|=r_{1} \\ z_{2}=r_{2}\left(\cos \theta_{2}+i \sin \theta_{2}\right) \\ \left|z_{2}\right|=r_{2} \\\\ z_1z_2=r_{1} r_{2}\left[\cos \left(\theta_{1}+\theta_{2}\right)+i \sin \left(\theta_{1}+\theta_{2}\right)\right] \\ =r_{1}\left(\cos \theta_{1}+i \sin \theta_{1}\right) \cdot r_{2}\left(\cos \theta_{2}+i \sin \theta_{2}\right) \\ \end{array}
\begin{array}{c} =r_{1} r_{2}\left[\left(\cos \theta_{1} \cos \theta_{2}-\sin \theta_{1} \sin \theta_{2}\right)+i\left(\sin \theta_{1} \cos \theta_{2}+\cos \theta_{1} \sin \theta_{2}\right)\right] \\ =r_{1} r_{2}\left[\cos \left(\theta_{1}+\theta_{2}\right)+i \sin \left(\theta_{1}+\theta_{2}\right)\right]\left|z_{1} z_{2}\right|=\left|z_{1} \| z_{2}\right| \end{array}

Hence, option a is correct.

Question:42

The point represented by the complex number 2 -i is rotated about origin through an angle \pi/2 in the clockwise direction, the new position of point is:
A. 1 + 2i
B. -1-2i
C. 2 + i
D. -1 + 2i

Answer:

The answer is the option (b).
If z rotated through an angle of \pi/2about the origin in clockwise direction .
Then the new position =z.e^{-\left ( \frac{\pi}{2} \right )}
=\left (2-i \right ).e^{-\left ( \frac{\pi}{2} \right )}
=\left (2-i \right ).\left [ \cos\left (- \frac{\pi}{2} \right )+isin\left (- \frac{\pi}{2} \right ) \right ]
=\left (2-i \right ).\left (0-i \right )
=-1-2i
Hence, the correct option is (b)

Question:43

Let x, y \in R, then x + iy is a non-real complex number if:
A. x = 0
B.y = 0
C. x \neq 0
D. y \neq 0

Answer:

The answer is the option (d).
x+iy is a non real complex number if y\neq0,

Hence, d is correct

Question:44

If a + ib = c + id, then
A. a^2 + c^2 = 0
B. b^2 + c^2 = 0
C. b^2 + d^2 = 0
D. a^2 + b^2 = c^2 + d^2

Answer:

The answer is the option (d).
a+ib=c+id
|a+ib|=|c+id|
\sqrt{a^2+b^2} =\sqrt{c^2+d^2}
Squaring both sides, a^2+b^2=c^2+d^2
Hence, d is correct

Question:45

The complex number z which satisfies the condition \left |\frac{i+z}{i-z} \right |=1 lies on
A. circle x^2 + y^2 = 1
B. the x-axis
C. the y-axis
D. the line x + y = 1

Answer:

The answer is the option (b).
\begin{array}{c} \text { Let } z=x+i y \quad\left|\frac{i+x+i y}{i-x-y i}\right|=1 \\\\ \left|\frac{x-(y+1) i}{-x-(y-1) i}\right|=1 \\\\ |x+(y+1) i|=|-x-(y-1) i| \\\\ \sqrt{x^{2}+(y+1)^{2}}=\sqrt{x^{2}+(y-1)^{2}} \\\\ x^{2}+(y+1)^{2}=x^{2}+(y-1)^{2} \\\\ y^{2}+2 y+1=y^{2}-2 y+1 \\ 2 y=-2 y \\\\ y=0 ; x-a x i s \end{array}

Hence, b is correct

Question:46

If z is a complex number, then
A.|z^2| > |z|^2
B. |z^2| \: \: equals\: \: |z|^2
C.|z^2| < |z|^2
D. |z^2| \geq |z|^2

Answer:

. The answer is the option (b).
Let \: \: z=x+iy \: \: \: \: ,|z|^2=x^2+y^2
Now,z^2=x^2+y^2 i^2+2xyi
z^2=x^2-y^2+2xyi
|z|^2=\sqrt{(x^2-y^2 )^2+(2xy)^2 }
=\sqrt{x^4+y^4+2x^2 y^2} =\sqrt{(x^2+y^2 )^2}
|z|^2=x^2+y^2=|z|^2

Hence, b is correct

Question:47

|z_1 + z_2| = |z_1| + |z_2| is possible if
A. z_1 =\bar{z_1}

B. z2=z1
C. arg (z_1) = arg (z_2)
D. |z_1| = |z_2|

Answer:

The answer is the option (c).
Since, |z_1+z_2 |=|z_1 |+|z_2 |
|z_1+z_2 |=|r_1 \left ( \cos \theta _1+i\sin\theta _1 \right )+r_2 \left ( \cos \theta _2+i\sin\theta _2 \right )|
=\sqrt{r_{1}^{2} \cos ^{2} \theta_{1}+r_{2}^{2} \cos ^{2} \theta_{2}+2 r_{1} r_{2} \cos \theta_{1} \cos \theta_{2}+r_{1}^{2} \sin ^{2} \theta_{1}+r_{2}^{2} \sin ^{2} \theta_{2}+2 r_{1} r_{2} \sin \theta_{1} \sin \theta_{2}}
So, \begin{array}{c} =\sqrt{r_{1}^{2}+r_{2}^{2}+2 r_{1} r_{2} \cos \left(\theta_{1}-\theta_{2}\right)} \\ \left|z_{1}+z_{2}\right|=\left|z_{1}\right|+\left|z_{2}\right| \\\\ \sqrt{r_{1}^{2}+r_{2}^{2}+2 r_{1} r_{2} \cos \left(\theta_{1}-\theta_{2}\right)}=r_{1}+r_{2} \end{array}
Squaring both sides,
\begin{array}{c} r_{1}^{2}+r_{2}^{2}+2 r_{1} r_{2} \cos \left(\theta_{1}-\theta_{2}\right)=r_{1}^{2}+r_{2}^{2}+2 r_{1} r_{2} \\ 2 r_{1} r_{2}-2 r_{1} r_{2} \cos \left(\theta_{1}-\theta_{2}\right)=0 \\ 1-\cos \left(\theta_{1}-\theta_{2}\right)=0 \\ \cos \left(\theta_{1}-\theta_{2}\right)=1 \\ \theta_{1}-\theta_{2}=0 \\ \theta_{1}=\theta_{2} \\ \operatorname{so} \arg \left(z_{1}\right)=\arg \left(z_{2}\right) \end{array}

Question:48

The real value of θ for which the expression \frac{1+i \cos \theta}{1-2 i \cos \theta} is a real number is:
A. n\pi+\frac{\pi}{4}


B. n\pi+(-1)^n\frac{\pi}{4}
C. 2n\pi+\frac{\pi}{2}
D. none of these

Answer:

Let \: \: z=\frac{1+i \cos \theta}{1-2 i \cos \theta} =\frac{1+i \cos \theta}{1-2 i \cos \theta} * \frac{1+2 i \cos \theta}{1+2 i \cos \theta} On solving we get
\frac{1+3 i \cos \theta-2 \cos ^{2} \theta}{1-4 i^{2} \cos ^{2} \theta} \\=\frac{1+3 i \cos \theta-2 \cos ^{2} \theta}{1+4 \cos ^{2} \theta}
=\frac{\left(1-2 \cos ^{2} \theta\right)+3 i \cos \theta}{1+4 \cos ^{2} \theta}
If z is real number then
\frac{3 \cos \theta}{1+4 \cos ^{2} \theta}=0
\\3 \cos \theta=0 \\\cos \theta=0
\theta=\frac{(2 n+1) \pi}{2}, n \in N

Question:49

The value of arg (x) when x < 0is:
A. 0
B. \frac{\pi}{2}

C. π
D. none of these

Answer:

The answer is the option (c).
\\ Let\: \: z= -x+0i \: \: and \: \: x<0 \\ |z|=\sqrt{(-1)^2+(0)^2 }=1,x<0

Since, the point (-x,0)lies on the negative side of the real axis , principal argument (z)=π

Hence, option c is correct

Question:50

If f(z)=\frac{7-z}{1-z^2}where z = 1 + 2i, then |f(z)| is
A. \frac{\left | z \right |}{2}


B. |z|
C. 2|z|
D. none of these

Answer:

The answer is the option (a).
Given that \\ z=1+2i \\ |z|=\sqrt{(1)^2+(2)^2 }=\sqrt5
Now f(z)=\frac{7-z}{1-z^2}
\begin{aligned} =\frac{7-(1+2 i)}{1-(1+2 i)^{2}} &=\frac{7-1-2 i}{1-1-4 i^{2}-4 i} \\ =\frac{6-2 i}{4-4 i} &=\frac{3-i}{2-2 i} \end{aligned}
\begin{aligned} =\frac{3-i}{2-2 i} * \frac{2+2 i}{2+2 i}=\frac{6+6 i-2 i-2 i^{2}}{4-4 i} \\ =\frac{6+4 i+2}{4+4} &=\frac{8+4 i}{8} \\ =1+\frac{1}{2} i & \end{aligned}
f(z)=\sqrt{(1)^{2}+\left(\frac{1}{2}\right)^{2}}=\sqrt{1+\frac{1}{4}}=\frac{\sqrt{5}}{2}=\frac{|z|}{2}

Hence, option a is correct

Important Notes from NCERT Exemplar Class 11 Maths Solution chapter 5

Various laws need to be followed and derived by a mathematician when it comes to multiplying the two complex numbers. All of them are named over the features of the specific formula derivation. Ahead, you will also learn about the Modulus and Conjugate of complex numbers and how to solve sums with their help. There is an Argand and Polar representation in the chapter, relating to graphical presentations. A point needs to be derived at the ‘x’ and ‘y’ axis to form an argand or complex plane.

Students, with the help of NCERT Exemplar Class 11 Maths solutions chapter 5, will not face any issues when trying to solve such problems. It will also help them score better in exams.

All the concepts have been covered in NCERT Exemplar solutions for Class 11 Maths chapter 5. By using NCERT Exemplar Class 11 Maths chapter 5 solutions PDF Download function, students can access quality study material that is effectively constructed by experts for the best learning experience.

Main Subtopics in NCERT Exemplar Class 11 Maths Solution Chapter 5

  • Introduction
  • Complex Numbers
  • Algebra of Complex Numbers
  • Addition of two complex numbers
  • Difference of two complex numbers
  • Multiplication of two complex numbers
  • Division of two complex number
  • Power of i
  • The square roots of a negative real number
  • Identities
  • The Modulus and the Conjugate of a Complex Number
  • Argand Plane and Polar Representation
  • polar representation of a complex number
  • Quadratic Equations

What will the students learn from NCERT Exemplar Class 11 Maths Solution Chapter 5?

Belonging to the algebraic part of the subject, it is always fun for the students to solve these problems, once they are clear with the concept. NCERT Exemplar Class 11 Maths chapter 5 solutions will help in understanding the said concept and give examples in how to apply them to questions of any kind.

Other than that, in geometric applications, complex numbers are also very useful, for students in their near future. Students, through NCERT Exemplar Class 11 Maths solutions chapter 5, will also learn graphs in an advanced manner with newer placements and some difficult problems, which would make them sharper and clever.

NCERT Solutions for Class 11 Mathematics Chapters

Important Topics in NCERT Exemplar Class 11 Maths Solutions Chapter 5

· Class 11 Maths NCERT Exemplar solutions chapter 5 has detailed that complex numbers, their operations, square root, power, identities, representation of complex numbers and quadratic equations are important topics which students should pay extra attention to.

· In NCERT Exemplar Class 11 Maths chapter 5 solutions, students will also learn about quadratic equations. A quadratic equation is an equation where rearrangement of a particular equation is done via a standard form, where x is the unknown number, whereas, a, b and c, are known numbers. An equation with four numbers is said to be quadratic equations.

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Yes, for those who want to score high in their school exams and also want to do better in entrance exams, practising NCERT questions is a must.

2. Are these solutions detailed?

Yes, our experts have solved every question in the most detail-oriented way in NCERT Exemplar Class 11 Maths solutions Chapter 5. 

3. Which topics are added in this chapter?

In this chapter, the complex numbers are explained, their properties, operations on complex numbers along with quadratic equations.

4. Are these solutions downloadable?

Yes, the solutions can be downloaded using the NCERT Exemplar Class 11 Maths Chapter 5 solutions PDF Download function, as one can get the offline PDF after clicking on the download link.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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